Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{r - 3}{r^2 + 3r - 18} \times \dfrac{-2r - 12}{-7r + 63} $
Explanation: First factor the quadratic. $q = \dfrac{r - 3}{(r + 6)(r - 3)} \times \dfrac{-2r - 12}{-7r + 63} $ Then factor out any other terms. $q = \dfrac{r - 3}{(r + 6)(r - 3)} \times \dfrac{-2(r + 6)}{-7(r - 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (r - 3) \times -2(r + 6) } { (r + 6)(r - 3) \times -7(r - 9) } $ $q = \dfrac{ -2(r - 3)(r + 6)}{ -7(r + 6)(r - 3)(r - 9)} $ Notice that $(r - 3)$ and $(r + 6)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -2(r - 3)\cancel{(r + 6)}}{ -7\cancel{(r + 6)}(r - 3)(r - 9)} $ We are dividing by $r + 6$ , so $r + 6 \neq 0$ Therefore, $r \neq -6$ $q = \dfrac{ -2\cancel{(r - 3)}\cancel{(r + 6)}}{ -7\cancel{(r + 6)}\cancel{(r - 3)}(r - 9)} $ We are dividing by $r - 3$ , so $r - 3 \neq 0$ Therefore, $r \neq 3$ $q = \dfrac{-2}{-7(r - 9)} $ $q = \dfrac{2}{7(r - 9)} ; \space r \neq -6 ; \space r \neq 3 $